✓ Board-verified
Mathematics · 2023 · 5 marks
CBSE 2023 · Region 5 · Set 1 · Q32
One observer estimates the angle of elevation to the basket of a hot air balloon to be $\displaystyle 60$°, while another observer $\displaystyle 100$ m away estimates the angle of elevation to be $\displaystyle 30^{\circ}$. Find :(a)The height of the basket from the ground.(b)The distance of the basket from the first observer's eye.(c)The horizontal distance of the second observer from the basket.
One observer estimates the angle of elevation to the basket of a hot air balloon to be $\displaystyle 60$°, while another observer $\displaystyle 100$ m away estimates the angle of elevation to be $\displaystyle 30^{\circ}$. Find :
(a)
The height of the basket from the ground.
(b)
The distance of the basket from the first observer's eye.
(c)
The horizontal distance of the second observer from the basket.
Marking-scheme solution
Let B is the basket of hot air balloon
\[\begin{aligned}
& \tan 60^{\circ}=\sqrt{3}=\frac{h}{x} \Rightarrow h=x \sqrt{3} \\
& \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{h}{x+100} \Rightarrow x=h \sqrt{3}-100
\end{aligned}
\]
using (i) and (ii)
(a) \(\displaystyle \mathbf{h}=(\mathbf{h} \sqrt{\mathbf{3}}-\mathbf{1 0 0}) \sqrt{\mathbf{3}}=\mathbf{3 h}-\mathbf{1 0 0} \sqrt{\mathbf{3}} \Rightarrow \mathbf{h}=\mathbf{5 0} \sqrt{\mathbf{3}} \mathbf{m}\)
(b) \(\displaystyle \sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{h}{y}=\frac{50 \sqrt{3}}{y} \Rightarrow \mathrm{y}=100 \mathrm{~m}\)
(c) \(\displaystyle \mathrm{x}=\frac{h}{\sqrt{3}}=50 \mathrm{~m} \Rightarrow \mathrm{x}+100=150 \mathrm{~m}\)
# ANOTHER SOLUTION AS PER BELOW FIGURE IS ALSO POSSIBLE
Let \(\displaystyle B\) is the basket of hot air balloon. \(\displaystyle D\) and \(\displaystyle C\) be the positions of the first and second observer's respectively.
\[\begin{aligned}
& \tan 60^{\circ}=\sqrt{3}=\frac{h}{x} \Rightarrow h=x \sqrt{3} \\
& \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{h}{100-x} \Rightarrow \sqrt{3} h=100-x
\end{aligned}
\]
(a)
using (i) and (ii)
\[\begin{aligned}
& h=\sqrt{3}(100-\sqrt{3} h) \Rightarrow h=25 \sqrt{3} \mathrm{~m} \\
& \text { (b) } \sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{h}{B D} \Rightarrow B D=50 \mathrm{~m} \\
& \text { (c) } \mathrm{x}=\frac{h}{\sqrt{3}}=25 \mathrm{~m} \\
& \Rightarrow \mathrm{AC}=100-\mathrm{x}=75 \mathrm{~m}
\end{aligned}
\]
More from Some Applications of Trigonometry
- From a point on the ground, which is 30 m away from the foot of a vertical tower, the angle of elevation of…2024 · asked 3×
- A wire is attached from a point A on the ground to the top of a pole BC, making an angle of elevation as…2026 · asked 3×
- Case Study-1: Kite festival is celebrated in many countries at different times of the year. In India, every…2022 · asked 3×
- Passenger boarding stairs, sometimes referred to as boarding ramps, stair cars or aircraft steps, provide a…2025 · asked 3×
- Find the length of the shadow on the ground of a pole of height 18 m when angle of elevation θ of the sun is…2023 · asked 3×
- Assertion: A ladder leaning against a wall, stands at a horizontal distance of 6 m from the wall. If the…2025 · asked 3×
- If a pole 6 m high casts a shadow 2 √3 m long on the ground, then sun's elevation is:2023 · asked 3×
- Gadisar Lake is located in the Jaisalmer district of Rajasthan. It was built by the King of Jaisalmer and…2022 · asked 3×
CBSE Class 10 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.